Electromagnetic Induction And Alternating Current Ques 94
- In the circuit shown, $L=1 \mu H, C=1 \mu F$ and $R=1 k \Omega$. They are connected in series with an AC source $V=V _0 \sin \omega t$ as shown. Which of the following options is/are correct?
(2017 Adv.)
(a) At $\omega \sim 0$, the current flowing through the circuit becomes nearly zero
(b) The frequency at which the current will be in phase with the voltage is independent of $R$
(c) The current will be in phase with the voltage if $\omega=10^{4} rads^{-1}$
(d) At $\omega \gg 10^{6} rads^{-1}$, the circuit behaves like a capacitor
Show Answer
Answer:
Correct Answer: 94.(a, b)
Solution:
Formula:
- At $\omega \approx 0, X _C=\frac{1}{\omega C}=\infty$. Therefore, current is nearly zero.
Further at resonance frequency, current and voltage are in phase. This resonance frequency is given by
$\omega _r=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{10^{-6} \times 10^{-6}}}=10^{6} rad / s$
We can see that this frequency is independent of $R$.
Further, $X _L=\omega L, X _C=\frac{1}{\omega C}$
At, $\omega=\omega _r=10^{6} rad / s, X _L=X _C$.
For $\omega>\omega _r, X _L>X _C$. So, circuit is inductive.
BETA
